Tính chất Ảnh_(toán_học)

Chung

Với mọi f : X → Y {\displaystyle f:X\rightarrow Y} các tập con A ⊆ X {\displaystyle A\subseteq X} , B ⊆ Y {\displaystyle B\subseteq Y} , ta có:

Hình ảnhTiền đề
f ( X ) ⊆ Y {\displaystyle f(X)\subseteq Y} f − 1 ( Y ) = X {\displaystyle f^{-1}(Y)=X}
f ( f − 1 ( Y ) ) = f ( X ) {\displaystyle f(f^{-1}(Y))=f(X)} f − 1 ( f ( X ) ) = X {\displaystyle f^{-1}(f(X))=X}
f ( f − 1 ( B ) ) ⊆ B {\displaystyle f(f^{-1}(B))\subseteq B}
(ta có dấu bằng nếu B ⊆ f ( X ) {\displaystyle B\subseteq f(X)} , ví dụ như nếu f {\displaystyle f} là một toàn ánh) [2][3]		 f − 1 ( f ( A ) ) ⊇ A {\displaystyle f^{-1}(f(A))\supseteq A}
(ta có dấu bằng bằng nếu f {\displaystyle f} là một đơn ánh)
f ( f − 1 ( B ) ) = B ∩ f ( X ) {\displaystyle f(f^{-1}(B))=B\cap f(X)} ( f | A ) − 1 ( B ) = A ∩ f − 1 ( B ) {\displaystyle (f\vert _{A})^{-1}(B)=A\cap f^{-1}(B)}
f ( f − 1 ( f ( A ) ) ) = f ( A ) {\displaystyle f(f^{-1}(f(A)))=f(A)} f − 1 ( f ( f − 1 ( B ) ) ) = f − 1 ( B ) {\displaystyle f^{-1}(f(f^{-1}(B)))=f^{-1}(B)}
f ( A ) = ∅ ⇔ A = ∅ {\displaystyle f(A)=\varnothing \Leftrightarrow A=\varnothing } f − 1 ( B ) = ∅ ⇔ B ⊆ Y ∖ f ( X ) {\displaystyle f^{-1}(B)=\varnothing \Leftrightarrow B\subseteq Y\setminus f(X)}
f ( A ) ⊇ B ⇔ ∃ C ⊆ A : f ( C ) = B {\displaystyle f(A)\supseteq B\Leftrightarrow \exists C\subseteq A:f(C)=B} f − 1 ( B ) ⊇ A ⇔ f ( A ) ⊆ B {\displaystyle f^{-1}(B)\supseteq A\Leftrightarrow f(A)\subseteq B}
f ( A ) ⊇ f ( X ∖ A ) ⇔ f ( A ) = f ( X ) {\displaystyle f(A)\supseteq f(X\setminus A)\Leftrightarrow f(A)=f(X)} f − 1 ( B ) ⊇ f − 1 ( Y ∖ B ) ⇔ f − 1 ( B ) = X {\displaystyle f^{-1}(B)\supseteq f^{-1}(Y\setminus B)\Leftrightarrow f^{-1}(B)=X}
f ( X ∖ A ) ⊇ f ( X ) ∖ f ( A ) {\displaystyle f(X\setminus A)\supseteq f(X)\setminus f(A)} f − 1 ( Y ∖ B ) = X ∖ f − 1 ( B ) {\displaystyle f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)}
f ( A ∪ f − 1 ( B ) ) ⊆ f ( A ) ∪ B {\displaystyle f(A\cup f^{-1}(B))\subseteq f(A)\cup B} [4] f − 1 ( f ( A ) ∪ B ) ⊇ A ∪ f − 1 ( B ) {\displaystyle f^{-1}(f(A)\cup B)\supseteq A\cup f^{-1}(B)}
f ( A ∩ f − 1 ( B ) ) = f ( A ) ∩ B {\displaystyle f(A\cap f^{-1}(B))=f(A)\cap B} f − 1 ( f ( A ) ∩ B ) ⊇ A ∩ f − 1 ( B ) {\displaystyle f^{-1}(f(A)\cap B)\supseteq A\cap f^{-1}(B)}
  • f ( A ) ∩ B = ∅ ⇔ A ∩ f − 1 ( B ) = ∅ {\displaystyle f(A)\cap B=\varnothing \Leftrightarrow A\cap f^{-1}(B)=\varnothing }

Nhiều hàm

Cho hai hàm f : X → Y {\displaystyle f:X\rightarrow Y} và g : Y → Z {\displaystyle g:Y\rightarrow Z} và các tập con A ⊆ X {\displaystyle A\subseteq X} , C ⊆ Z {\displaystyle C\subseteq Z} , ta có:

  • ( g ∘ f ) ( A ) = g ( f ( A ) ) {\displaystyle (g\circ f)(A)=g(f(A))}
  • ( g ∘ f ) − 1 ( C ) = f − 1 ( g − 1 ( C ) ) {\displaystyle (g\circ f)^{-1}(C)=f^{-1}(g^{-1}(C))}

Nhiều tập hợp

Cho hàm f : X → Y {\displaystyle f:X\rightarrow Y} và các tập con A 1 , A 2 ⊆ X {\displaystyle A_{1},A_{2}\subseteq X} , B 1 , B 2 ⊆ Y {\displaystyle B_{1},B_{2}\subseteq Y} , ta có:

Hình ảnhTiền đề
A 1 ⊆ A 2 ⇒ f ( A 1 ) ⊆ f ( A 2 ) {\displaystyle A_{1}\subseteq A_{2}\Rightarrow f(A_{1})\subseteq f(A_{2})} B 1 ⊆ B 2 ⇒ f − 1 ( B 1 ) ⊆ f − 1 ( B 2 ) {\displaystyle B_{1}\subseteq B_{2}\Rightarrow f^{-1}(B_{1})\subseteq f^{-1}(B_{2})}
f ( A 1 ∪ A 2 ) = f ( A 1 ) ∪ f ( A 2 ) {\displaystyle f(A_{1}\cup A_{2})=f(A_{1})\cup f(A_{2})} [4][5] f − 1 ( B 1 ∪ B 2 ) = f − 1 ( B 1 ) ∪ f − 1 ( B 2 ) {\displaystyle f^{-1}(B_{1}\cup B_{2})=f^{-1}(B_{1})\cup f^{-1}(B_{2})}
f ( A 1 ∩ A 2 ) ⊆ f ( A 1 ) ∩ f ( A 2 ) {\displaystyle f(A_{1}\cap A_{2})\subseteq f(A_{1})\cap f(A_{2})}
(ta có dấu bằng nếu f {\displaystyle f} là đơn ánh [6])		 f − 1 ( B 1 ∩ B 2 ) = f − 1 ( B 1 ) ∩ f − 1 ( B 2 ) {\displaystyle f^{-1}(B_{1}\cap B_{2})=f^{-1}(B_{1})\cap f^{-1}(B_{2})}
f ( A 1 ∖ A 2 ) ⊇ f ( A 1 ) ∖ f ( A 2 ) {\displaystyle f(A_{1}\setminus A_{2})\supseteq f(A_{1})\setminus f(A_{2})} (ta có dấu bằng nếu f {\displaystyle f} là đơn ánh) f − 1 ( B 1 ∖ B 2 ) = f − 1 ( B 1 ) ∖ f − 1 ( B 2 ) {\displaystyle f^{-1}(B_{1}\setminus B_{2})=f^{-1}(B_{1})\setminus f^{-1}(B_{2})}
f ( A 1 △ A 2 ) ⊇ f ( A 1 ) △ f ( A 2 ) {\displaystyle f(A_{1}\triangle A_{2})\supseteq f(A_{1})\triangle f(A_{2})}
(ta có dấu bằng nếu f {\displaystyle f} là đơn ánh)		 f − 1 ( B 1 △ B 2 ) = f − 1 ( B 1 ) △ f − 1 ( B 2 ) {\displaystyle f^{-1}(B_{1}\triangle B_{2})=f^{-1}(B_{1})\triangle f^{-1}(B_{2})}

Ngoài ra

  • f ( ⋃ s ∈ S A s ) = ⋃ s ∈ S f ( A s ) {\displaystyle f\left(\bigcup _{s\in S}A_{s}\right)=\bigcup _{s\in S}f(A_{s})}
  • f ( ⋂ s ∈ S A s ) ⊆ ⋂ s ∈ S f ( A s ) {\displaystyle f\left(\bigcap _{s\in S}A_{s}\right)\subseteq \bigcap _{s\in S}f(A_{s})}
  • f − 1 ( ⋃ s ∈ S B s ) = ⋃ s ∈ S f − 1 ( B s ) {\displaystyle f^{-1}\left(\bigcup _{s\in S}B_{s}\right)=\bigcup _{s\in S}f^{-1}(B_{s})}
  • f − 1 ( ⋂ s ∈ S B s ) = ⋂ s ∈ S f − 1 ( B s ) {\displaystyle f^{-1}\left(\bigcap _{s\in S}B_{s}\right)=\bigcap _{s\in S}f^{-1}(B_{s})}